Question 1197158: An extract of a study conducted about the effects of a special class designed to aid students with
verbal skills is given below:
Each child was given a verbal skill test twice, both before and after completing a 4-week period
in the class. Let Y = score on exam at time 2 - score on exam at time 1. Hence, if the population
mean µ for Y is equal to 0, the class has no effect, on the average. For the four children in the
study, the observed values of Y are 8-5=3, 10-3=7, 5-2=3, and 7-4=3 (e.g. for the first child, the
scores were 5 on exam 1 and 8 on exam 2, so Y = 8-5=3). It is planned to test the null hypothesis
of no effect against the alternative hypothesis that the effect is positive, based on the following
results from a statistical software package:
Variable Number of cases Mean Standard Deviation Standard Error
Y 4 4.000 2.000 1.000
1.1.1 Set up the null and alternative hypotheses. (4)
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website!
Hi
Hi
Let Y = score on exam at time 2 - score on exam at time 1.
Y = {3,7,3,3} µ = 4 and σ = 2 SE = 1
Ho: Y = 0 | Ho must include =, ≤ , 0r ≥
Ha: Y > 0 Claim
'one-sided'
 = -4
At 95% Confidence Level, critical value = 3.122 | =CONFIDENCE(alpha,standard_dev,size) = =CONFIDENCE(.05,2,4) = 3.182
If the absolute value of the t-value is greater than the critical value, you reject the null hypothesis.
|-4| > 3.122 Reject Ho
The test results support the claim that the Special class had a positive effect
Wish You the Best in your Studies.
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